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Math Help - The square/cube/quartic... of a binomial/trinomial/multinomial?

  1. #1
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    The square/cube/quartic... of a binomial/trinomial/multinomial?

    Is there a general rule for finding (a+b)^n/(a+b+c)^n/(a+b+c+...)^n?

    For all I know:

    (a+b)^1 = a^1+b^1 = a+b

    (a+b)^2 = a^2+2ab+b^2

    (a+b)^3 = a^3+3a^2b+3ab^2+b^3

    [..................................................  .........]<br />


    (a+b+c)^1 = a^1+b^1+c^1 = a+b+c

    (a+b+c)^2 = (a+b+c)(a+b+c) = a(a+b+c)+c(a+b+c)+c(a+b+c) = a^2+ab+ac+ac+bc+c^2+ac+bc+c^2 = a^2+b^2+c^2+ab+ab+ac+ac+bc+bc = a^2+b^2+c^2+2ab+2ac+2bc = a^2+b^2+c^2+2(ab+ac+bc)
    [..........................................]

    But after that, it just gets tedious!
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  2. #2
    Super Member bigwave's Avatar
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    (a+b+c)^1 = a^1+b^1+c^1 = a+b+c

    this expression is true only because the exponent is 1
    if it were another power you could not do this
    other wise the rule is power to power
     <br /> <br />
(xy)^\alpha \ = \ x^\alpha y^\alpha<br />

    as to (a + b)^\alpha ussually if \alpha>3 you break it into mutliple expressions using the rule x^\alpha x^\beta = x^{\alpha + \beta}
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  3. #3
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    This should help you .
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  4. #4
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    Quote Originally Posted by Viral View Post
    This should help you .
    Well, I've no problem with solving polynomials. I'm just looking for a more general rule for expansions. For example, if I want to find (a+b+c+d)^3, am I supposed to calculate (a+b+c+d)(a+b+c+d)(a+b+c+d)? I mean, shouldn't there be an easier way to find this?
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  5. #5
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    Quote Originally Posted by bigwave View Post
    (a+b+c)^1 = a^1+b^1+c^1 = a+b+c

    this expression is true only because the exponent is 1
    if it were another power you could not do this
    other wise the rule is power to power
    You are right.

     <br />
(xy)^\alpha \ = \ x^\alpha y^\alpha<br />

    as to (a + b)^\alpha ussually if \alpha>3 you break it into mutliple expressions using the rule x^\alpha x^\beta = x^{\alpha + \beta}
    I don't understand. If, say, I want to find (a+b)^4, I need to break it to [(a+b)^2][(a+b)^2] = (a^2+2ab+b^2)(a^2+2ab+b^2)? That I understand. But I don't understand how using x^{\alpha}x^{\beta} = x^{\alpha+\beta} is going to help me calculate (a+b)^4.
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  6. #6
    Super Member bigwave's Avatar
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    the exponent rules ussually just use x as default

    x^2 can mean (.......)^2

    x^{\alpha}x^{\beta} = x^{\alpha+\beta}
    as you mention probably doesn't relate to much here
    but in certain other expressions it would

    what you are doing is correct...
    it can just get very busy....
    on the TI89 use the expand((a+b)^4) and its done

    calculus has some better ways to deal with this.

    aloha and have nice weekend
    Last edited by bigwave; November 6th 2009 at 05:41 PM.
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